4

I am aware of companion matrices for single variable polynomials:

$$p(x) = x^k+c_{k-1}x^{k-1}+\cdots +c_0$$

$${\bf C_p} = \left[\begin{array}{ll}{\bf 0}^T & -c_0 \\ {\bf I}_{k-1} & {\bf c}_{1:k}\end{array}\right]$$

$\bf C_p$ will have the same eigenvalues as the roots of $p(x)=0$

Are there any ways to build companion matrices for multivariate polynomials (in some sense)?

EDIT for example what I would like is that if I have $f(x,y) = (x+1)^2+(y+1)^2-1$ I can (by some recipe or algorithm) build a matrix that has the same properties in the sense of approximating roots of the polynomial when iterating matrix multiplication in the same way as one variable do.

Rodrigo de Azevedo
  • 23,223
mathreadler
  • 26,534
  • Interesting question! For example, given a bivariate polynomial $f(x,y)$, you seek a "formula" which yields $n{,\times,}n$ matrices $A,B$ (for some positive integers $n$, and to be useful, you probably also want $A,B$ to commute) such that $f(A,B)=0$. Yes? – quasi May 31 '17 at 14:31
  • I was thinking one big matrix for both x,y simultaneously somehow, but maybe one per variable could work. – mathreadler May 31 '17 at 14:32
  • If you want just one matrix for $f(x,y)$, wouldn't the companion matrix of $f(x,x)$ work? – quasi May 31 '17 at 14:36
  • But how to "expand" the matrix to contain two variables? – mathreadler May 31 '17 at 15:40
  • I don't follow. Can you give a specific example? – quasi May 31 '17 at 15:41
  • $f(x,y) = (x+1)^2+(y+1)^2-1$ say I want to construct a matrix which has the same properties as companion matrices for 1 variable in the sense that we can iterate matrix multiplication to estimate solutions to $f(x,y)=0$ – mathreadler May 31 '17 at 15:43
  • There are two problems with this. First how do you define the characteristic polynomial of a matrix in two variables: $\det((x + y)I - A)$, $\det(xyI - A)$, something else? Second, keep in mind that $k[x,y]$ is not a PID so that structure theorem goes out the window. – Sera Gunn May 31 '17 at 16:14
  • What @quasi suggests with two commuting matrices seems like the most natural way to answer this question. – Sera Gunn May 31 '17 at 16:18
  • @quasi Feel free to write an answer based on the commuting matrices even if it's not what I had in mind. – mathreadler Jun 03 '17 at 16:00
  • 1
    @mathreadler: Thanks, but I don't think I added enough to warrant an answer. The interesting question relating to the two-variable idea, but which I haven't really thought about is, given a positive integer $n$ and a polynomial $f(x,y)$, how to characterize the pairs $(A,B)$ of commuting $n{,\times,}n$ matrices such that $f(A,B)=0$. – quasi Jun 03 '17 at 16:11
  • @mathreadler: Maybe ask a new question (once you figure out what the right question is) with a link back to this one. – quasi Jun 03 '17 at 16:18
  • Ok, I thought that maybe you had an answer but were reluctant to share in case it would not be what I hoped. @quasi yes that's what I usually try to do, make a new better question and link it. – mathreadler Jun 03 '17 at 16:24
  • 2
    see https://doi.org/10.1006/jcom.1999.0530 Multivariate Polynomials, Duality, and Structured Matrices They have some examples you are interested – mkatkov Jul 17 '18 at 20:13
  • @mkatkov Thank you. Wow it's very big, but I will try to read it. Do you happen to know how to make a matrix according to the article for the circle equation example? – mathreadler Jul 17 '18 at 20:27
  • Related: https://math.stackexchange.com/q/1630316/339790 – Rodrigo de Azevedo Jan 25 '20 at 13:14
  • One interesting property of companion matrices is that they are invariant under permutations. I like to think of them as building blocks for a canonical form (the Frobenius normal form, which is also called the rational normal form) under matrix conjugation. However, you seem to be interested in them from another point of view. – Malkoun Jan 25 '20 at 16:37
  • @Malkoun Aha. I have not considered this. It could be interesting. I am originally an electrical engineer and mostly try to find uses for mathematical things which I learn about in practical computational contexts for algorithms in various fields. I have heard about Frobenius form, but I have not investigated it so much. Thank you for the idea! – mathreadler Jan 25 '20 at 21:16

0 Answers0